two brownian motions in $ \mathbb{Z}^2 $

Question

I was wondering what is the probability for 2 brownian walkers coming from 2 different initial positions to be at the same position at time t.

I consider that at each step, each point can independently move of $(0,\pm1)$ or $(\pm1, 0)$ with probability $\frac14$ for each direction.

Answer 1

If the difference between the two starting positions is $(x,y)$, then the two particles are at the same position at time $n$ with probability $$ \frac1{4^{n+1}\pi^2}\iint_{[-\pi,\pi]^2}\left(\cos u+\cos v\right)^{2n}\,\cos(ux+vy)\,\mathrm du\mathrm dv. $$ To see this, note that a random variable $(X,Y)$ with values in $\mathbb Z^2$ equals $(0,0)$ with probability $$ \frac1{4\pi^2}\iint_{[-\pi,\pi]^2}E\left(\mathrm e^{\mathrm i(uX+vY)}\right)\,\mathrm du\mathrm dv. $$ Of course, whether the formula given in this post is explicit or not is at best debatable...

Answer 2

Brownian motion on $\mathbb{Z}^2$ is just a random walk. The motion in the $x$ and $y$ directions are independent walks.

In the x direction walk with probability $\frac{1}{4}$ of moving left-right, $\frac{1}{2}$ of staying in place.

You can draw something like Pascal's triangle in this case. In fact, just the even rows.

    1
  1 2 1 
1 4 6 4 1

Every time we step, we are asking if the coin is heads or tails.

$$ X_i = \left\{\begin{array}{cc} 1 & \mathbb{P}(H) = \frac{1}{2}\\ 0 & \mathbb{P}(T) = \frac{1}{2}\end{array} \right.$$

The random walk in the x axis is $Y = \sum_{i=0}^N X_i$.

For two random walkers to be in the same place, their difference must be zero. However, as a probability distribution $Y = - Y$.

$$ \mathbb{P}(Y_1 - Y_2 = 0) = \mathbb{P}(Y_1 + Y_2 = 0) = \mathbb{P}\left( \sum_{i=1}^{2N} = 0\right) = \binom{2n}{n}$$

For since both axes are independent, we take the square of this number $ \frac{1}{16^n}\binom{2n}{n}^2$